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1–14.

Evaluate each of the following to three significant figures and

express each answer in Sl units using an appropriate prefix:
(a) 354 mg(45 km) > (0.0356 kN), (b) (0.004 53 Mg) (201 ms),
and (c) 435 MN> 23.2 mm.

SOLUTION
C 354 A 10-3 B g D C 45 A 103 B m D
a) (354 mg)(45 km)>(0.0356 kN) =
0.0356 A 103 B N

0.447 A 103 B g # m
=
N
= 0.447 kg # m>N Ans.

b) (0.00453 Mg)(201 ms) = C 4.53 A 10-3 B A 103 B kg D C 201 A 10-3 B s D

= 0.911 kg # s Ans.

435 A 106 B N 18.75 A 109 B N

c) 435 MN>23.2 mm = = = 18.8 GN>m Ans.
23.2 A 10-3 B m m

Ans:
0.447 kg # m>N
0.911 kg # s
18.8 GN>m

14

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1–19.

If a man weighs 690 newtons on earth, specify (a) his mass

in kilograms. If the man is on the moon, where the
acceleration due to gravity is gm = 1.61 m>s2, determine
(b) his weight in newtons, and (c) his mass in kilograms.

Solution
690
(a) m = = 70.3 kg Ans.
9.81
1.61
(b) W = 690 c d = 113 N Ans.
9.81
690
(c) m = = 70.3 kg Ans.
9.81

Ans:
70.3 kg
113 N
70.3 kg

19

0133915441_ISM01-229791.indd 19 27/01/16 2:42 PM

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*2–12. The device is used for surgical replacement of the y¿ y

knee joint. If the force acting along the leg is 360 N,
determine its components along the x ¿ and y axes. 10⬚

x¿
x

60⬚

Solution
360 N
-Fx′ 360
= ; Fx′ = - 183 N Ans.
sin 30° sin 80°
Fy 360
= ; Fy = 344 N Ans.
sin 70° sin 80°

Ans:
Fx′ = - 183 N
Fy = 344 N

33
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2–17.

Two forces act on the screw eye. If F1 = 400 N and F1

F2 = 600 N, determine the angle u(0° … u … 180°)
between them, so that the resultant force has a magnitude
of FR = 800 N.

F2
SOLUTION
The parallelogram law of addition and triangular rule are shown in Figs. a and b,
respectively. Applying law of cosines to Fig. b,

800 = 24002 + 6002 - 2(400)(600) cos (180° - u°)

8002 = 4002 + 6002 - 480000 cos (180° - u)

cos (180° - u) = - 0.25

180° - u = 104.48

u = 75.52° = 75.5° Ans.

Ans:
u = 75.5°

38
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2–18.

Two forces F1 and F2 act on the screw eye. If their lines of F1

action are at an angle u apart and the magnitude of each
force is F1 = F2 = F, determine the magnitude of the
resultant force FR and the angle between FR and F1.

SOLUTION
F2
F F
=
sin f sin (u - f)

sin (u - f) = sin f

u - f = f

u
f = Ans.
2

FR = 2(F)2 + (F)2 - 2(F)(F) cos (180° - u)

Since cos (180° - u) = -cos u

FR = F A 22 B 21 + cos u

u 1 + cos u
Since cos a b =
2 A 2

Then

u
FR = 2F cosa b Ans.
2

Ans:
u
f =
2
u
FR = 2F cos a b
2

39
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2–26.

If FB = 3 kN and u = 45°, determine the magnitude of the y

resultant force of the two tugboats and its direction A
measured clockwise from the positive x axis.
FA 2 kN

30
x
u
C
SOLUTION FB

The parallelogram law of addition and the triangular rule are shown in Figs. a and b,
respectively.
B
Applying the law of cosines to Fig. b,

FR = 222 + 32 - 2(2)(3) cos 105°

= 4.013 kN = 4.01 kN Ans.

Using this result and applying the law of sines to Fig. b, yields

sin a sin 105°

= a = 46.22°
3 4.013

Thus, the direction angle f of FR, measured clockwise from the positive x axis, is

f = a - 30° = 46.22° - 30° = 16.2° Ans.

Ans:
FR = 4.01 kN
f = 16.2°

47
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3–3.

The members of a truss are pin connected at joint O. y

Determine the magnitude of F1 and its angle u for 5 kN
equilibrium. Set F2 = 6 kN.
70 F2
30
x
O u

SOLUTION 5
3
4
:
+ ©F = 0; 4 F1
x 6 sin 70° + F1 cos u - 5 cos 30° - (7) = 0 7 kN
5
F1 cos u = 4.2920
3
+ c ©Fy = 0; 6 cos 70° + 5 sin 30° - F1 sin u - (7) = 0
5
F1 sin u = 0.3521

Solving:

u = 4.69° Ans.

F1 = 4.31 kN Ans.

Ans:
u = 4.69°
F1 = 4.31 kN

165
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3–13.

The unstretched length of spring AB is 3 m. If the block is 3m 4m

held in the equilibrium position shown, determine the mass
of the block at D.
C B

kAC 20 N/m
3m
kAB 30 N/m

SOLUTION
A
F = kx = 30(5 - 3) = 60 N

:
+ ©F = 0; 4
x Tcos 45° - 60 a b = 0
5
T = 67.88 N
D
3
+ c ©Fy = 0; -W + 67.88 sin 45° + 60 a b = 0
5
W = 84 N
84
m = = 8.56 kg Ans.
9.81

Ans:
m = 8.56 kg

175
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3–21.

If the spring DB has an unstretched length of 2 m, determine

2m 3m
the stiffness of the spring to hold the 40-kg crate in the
position shown.
C B

2m
k

Solution
A
Equations of Equilibrium. Referring to the FBD shown in Fig. a,

+ ΣFx = 0;   TBD a 3 b - TCDa 1 b = 0

S (1)
113 12

2 1
+cΣFy = 0;      TBD a b + TCDa b - 40(9.81) = 0 (2)
113 12
Solving Eqs (1) and (2)

TBD = 282.96 N   TCD = 332.96 N

The stretched length of the spring is

l = 232 + 22 = 213 m

Then, x = l - l0 = ( 113 - 2 ) m. Thus,

Fsp = kx; 282.96 = k ( 113 - 2 )
k = 176.24 N>m = 176 N>m Ans.

Ans:
k = 176 N>m

183
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3–35.

Each cord can sustain a maximum tension of 500 N.

Determine the largest mass of pipe that can be supported. D
5
3
4
C
B

60°
A E

SOLUTION
H
At H:

+ c ©Fy = 0; FHA = W

At A:

+ c ©Fy = 0; FAB sin 60° - W = 0

FAB = 1.1547 W

:
+ ©F = 0;
x FAE - (1.1547 W) cos 60° = 0

FAE = 0.5774 W

At B:

3
+ c ©Fy = 0; FBD a b - (1.1547 cos 30°)W = 0
5

FBD = 1.667 W

:
+ ©F = 0; 4
x -FBC + 1.667 W a b + 1.1547 sin 30° = 0
5
FBC = 1.9107 W

By comparison, cord BC carries the largest load. Thus

500 = 1.9107 W

W = 261.69 N

261.69
m = = 26.7 kg Ans.
9.81

Ans:
m = 26.7 kg

197